nonlinear macromodeling of amplifiers and applications to ...s-sanchez/622 op amp non-linera...

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Nonlinear Macromodeling of Amplifiers and

Applications to Filter Design.

Thanks to Heng Zhang for part of the material

ECEN 622

By Edgar Sánchez-Sinencio

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http://www.national.com/analog/amplifiers/spice_models

OP AMP MACROMODELS

http://www.analog.com/static/imported-files/application_notes/48136144500269408631801016AN138.pdf

Systems containing a significant number of Op Amps can take a lot of time of simulation when Op Amps are described at the transistor level. For instance a 5th order filter might involve 7 Op Amps and if each Op Amps contains say 12 to 15 transistors, the SPICE analysis of a circuit containing 60 to 75 Transistors can be too long and tricky in particular for time domain simulations. Therefore the use of a macromodel representing the Op Amp behavior reduces the simulation time and the complexity of the analysis. The simplicity of the analysis of Op Amps containing macromodels is because macromodels can be implemented using SPICE primitive components. Some examples of macromodels are discussed next.

ECEN 622(ESS) Analog and Mixed Signal Center TAMU

FUNDAMENTAL ON MACROMODELING USING ONLY PRIMITIVE SPICE COMPONENTS

1. Low Pass First Order

p1LP s1

kHω+

=

Option 1

1kRC1

p

=

=ω

Option 2

3 RC1RRk

10~A

p

1

9

=ω

=

inV oVR

C

CR

oVinVR1

∞→oA

C

R

R1

VX

Vin AVX

Vo

ECEN 622 (ESS) Analog and Mixed Signal Center TAMU

4

Option 3

1

inRV

VX

R1 gmVXR C

Vo

RC1;Rgk pm =ω=

2. Higher Order Low Pass

( )( )2p1p2LP s1s1

KHω+ω+

=

KK

CR1

CR1

222p

111p

=

=ω

=ω

FirstOrder

1pωVin VX

Vin R1

VXC1

FirstOrder

2pωKVX

Concept.

R2

C2

Vo

XVK ⋅

−

Note.- If you need to isolate the output use

a final VCVS with a gain of one

Le us consider a second-order case:

5

2o

o2o

3LPs

Qs

KHω+

ω+

=

( )

LR

Q

LC1

10H;LC1K

o

2o

3LPo

=ω

=ω

==

Resonator (one zero, two complex poles)

( )2o

o2z

Rs

Qs

s1kHω+

ω+

ω+=

RC1Q

LC1

LRLC1k

o

2o

z

=ω

=ω

=ω=

R

-Vin C

+

-Vo

-

++

-

R

CL

Vo

VoVin

ECEN 622 (ESS) Analog and Mixed Signal Center TAMU

Active RC Filter Design with Nonlinear Opamp Macromodel Design a two stage Miller CMOS Op Amp in

0.35 μm and propose a macromodel containing up to the seventh-harmonic component

Compare actual transistor model versus the proposed non-linear macromodel

Use both macromodel and transistor level to design a LP filter with H(o) =10dB, f3dB=5 MHz

Result comparison

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1st order Active-RC LP filter

7 ECEN 622(ESS) Analog and Mixed Signal Center TAMU

Filter transfer function with Ideal Opamp

8

2,

1 2

1( )(1 )LP ideal

RH sR sR C

= −+

H(o) = 10dB 2

1

RR

= 10dB = 3.16

f3dB=5 MHz 2

1R C = 6.28*5M = 31.4Mrad

(1)

(2)

(3)

Choose R1, R2 and C from equations (1) ~ (3). To minimize loading effect, R2 should be large enough. Here we choose R2 = 31.6kΩ, R1 = 10kΩ, and C = 1pF.

Filter transfer function with finite Opamp gain and GBW One pole approximation for Opamp Modeling: Av =

GB/s.(it holds when GBW >> f3dB and Av(0) >>1)

A two stage Miller Op amp is designed. GBW is chosen ~20 times the f3dB to minimize the finite GBW effect; GBW = 100MHz is also easy to achieve in 0.35μm CMOS technology. 9

2,

1 2 2

( )(1 )(1 )

LP nonidealRH s s sR sR C R

GB GB

−=

+ + +

Non-Linear Macro-Model for a source-degenerated OTA

ECEN 622(ESS) Analog and Mixed Signal Center TAMU

Linear Transistor Model:

G

D

S connected to Bulk

G D

S connected to Bulk

Cgs

Cgd

Rdsgm vgs Cdb

Linear relation

PoleHigh frequency Zero

Input capacitance

Linear OTA model:

Vin-

IO- IO+

Vin+

IBias

Non-Linear OTA model:

Vin-

I1 I2

Vin+

IDC

Let:

We can easily get:

Which can be expanded to:

To determine Odd Harmonic effects for an ideal OTA !!

How to Extract the Coefficients:

• By Sweeping the input voltage and integrating the output

current, we can these coefficients.

Generally if we have:

We can extract the coefficients by differentiation, where:

• a2 is ideally zero.

• Getting the first 3 coefficients only is a valid approximation.

A source degenerated OTA as an example:

OTA

Output current of one branch versus input differential voltage.

1st derivative 2nd derivative 3rd derivative

Coefficients:

a0=206.777 µA

a1=1.69094mA/v

a2=9.07µA/v2

a3= - 1.764mA/v3

The accuracy of these numbers depends on the number of points used in the

DC sweep. By taking more points, even harmonics reduce to zero.

Macromodel used:

1. Non-linear transfer function.

2. non-dominant pole .

3. Feed-forward path leads to Right half plane zero. (Cgd of the driver trans.)

4. Output Resistance and Load Capacitance.

(1)

(2)

(3)

(4)

DC sweep of Macromodel:

Changes due to measurement accuracy and number of points

AC response comparison:

Transistor level Macro-model

Two stage Miller Amplifier Design

19

Opamp Design parameters

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Power 278uA @ 3V

1st Stage PMOS(W/L) 30u/0.4u

NMOS(W/L) 15u/0.4u

2nd Stage PMOS(W/L) 120u/0.4u

NMOS(W/L) 60u/0.4u

Miller Compensation

Cm 800fF

Rm 400 Ω

OPAMP Frequency response

DC Gain: 53 dB, GBW: 86.6 MHz, phase margin: 69.7 deg. Dominant pole:154KHz, Second pole: 197MHz

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Output Spectrum of Open loop OPAMP

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1mVpp input @ 1KHz (THD= -49.2dB)

a1~a7 can be extracted from PSS simulation results:

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2 3 4 5 6 71 2 3 4 5 6 7out o d d d d d d dV a a v a v a v a v a v a v a v= + + + + + + +

a1 = DC gain = 450

22

1

41.922a AHD dB

a= = a2 = 2934

23

31

59.14

a AHD dBa

= = a3 = 2.16e6

Similarly, we can obtain: a4 = 6e7, a5 = 5.6e11, a6 = 1e13, a7 = 7e16

Opamp Macro model

Modeled: input capacitance, two poles, one RHP zero, nonlinearity, finite output resistance, and capacitance

Nonlinearity model should be placed before the poles to avoid poles multiplication

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Nonlinearity Model

uses mixer blocks to generate nonlinear terms model up to 7th order non-linearity set each VCCS Gain as the nonlinear coefficients. set the gain for 1st VCCS = gm1 = 512uA/V, gain for 2nd VCCS = gm2 =

2.85mA/V, and scale all the nonlinear coefficients derived above by a1.

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Opamp AC response: Transistor-level vs. Macromodel

Macro-model mimic the transistor level very well at frequencies below 10MHz discrepancy at higher frequency due to the higher order poles and zeros not modeled

in the Macromodel 26

Filter AC response: Transistor-level vs. Macromodel

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Output Spectrum (0dBm input @ 1KHz)

Macromodel(THD=-63dB)

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Transistor Level (THD = -66.4dB)

Performance Comparison Transistor Level Macro-model

-3dB BW 154KHz 180KHz GBW 86.6MHz 90MHz

DC Gain 53 dB 51.3 dB Phase Margin 69.7 degree 74.9 degree

THD: -50dBm @ 1KHz -49.2 dB -49.6 dB

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Table I. Open loop Opamp Performance Comparison

Table II. LPF Performance Comparison

Transistor Level Macro-model BW of LPF 4.9MHz 4.86MHz

DC Gain of LPF 9.95 dB 10.19 dB THD: 0dBm @ 1KHz -66.4 dB -63dB

Observation THD of the LPF at 0dBm input is better than that of the

open loop Opamp with a small input at -50dBm. This is because OPAMP gain is ~50 dB, when configured as a LPF, OPAMP input is attenuated by the feedback loop better linearity.

when keep increasing the input amplitude, the THD of the transistor-level degrades dramatically. This is because large swing activates more nonlinearity and even cause transistors operating out of saturation region; however, the THD of Macro-model doesn’t reflect this because we didn’t implement the limiter block.

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Gm-C Filter Design with Nonlinear Opamp Macromodel Use a three current mirror Transconductance

Amplifier. Compare actual transistor model versus the

non-linear macromodel Use both macromodel and transistor level to

design a LP filter with H(o) =10dB, f3dB=5 MHz Result Comparison

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1st order Gm-C LP filter

32 ECEN 622(ESS) Analog and Mixed Signal Center TAMU

Filter transfer function With Ideal OTA:

33

H(o) = 10dB 1

2

m

m

gg

= 10dB = 3.16

f3dB=5 MHz 2mgC

= 6.28*5M = 31.4Mrad

Output resistance of gm1 should be >> 1/gm2 Choose C = 4pF, gm2 = 0.126mA/V, gm1 = 0.4mA/V

1

2 2

1( )1 /

mideal

m m

gH sg sC g

= −+

Three Current mirrors OTA Design

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OTA Design parameters

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Power 240uA @ + 1.5V

Input NMOS 4u/0.6u

PMOS current mirror 12u/0.4u

NMOS current mirror 1u/0.4u

AC simulation of Gm: Transistor Level

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gm = 0.4mA/V, which is our desired value

its frequency response is good enough for a LPF with 5MHz cutoff frequency

OTA Output resistance: Transistor Level

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output resistance of the OTA >>1/gm2

Gm-C LPF Output spectrum: Transistor level

THD = -26dB for 0dBm input@1kHz 38

OTA Macro model

Since the internal poles and zeros are at much higher frequency than 5MHz, only the important ones are included in the macro-model

Nonlinearity model is the same as the Opamp in Active-RC filter

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AC simulation of Gm: Macro-model

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OTA Output resistance: Macro-model

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output resistance of the OTA >>1/gm2

Gm-C LPF Frequency response

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Gm-C LPF Output spectrum: Macromodel

THD = -33dB for 0dBm input@1kHz 43

Performance Comparison

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Table I. Gm-C Filter Performance Comparison

Table II. Comparison between Transistor Level Active-RC and Gm-C LPF

Transistor Level Macro-model Gm 409uA/V 412uA/V

BW of LPF 5.05MHz 5.05MHz DC Gain of LPF 10 dB 10 dB

THD: 0dBm @ 1KHz -26 dB -33dB

Active RC Gm-C DC gain 9.95dB 10dB

BW 4.9MHz 5.05MHz THD: 0dBm @ 1KHz -66.4 dB -26 dB

Noise Level 0.048µV/ @1kHz 0.05µV/ @1kHz Power 0.83mW 0.72mW

HzHz

Discussion With comparable DC gain, BW, Noise level and

Power consumption, Gm-C filter has much worse linearity than Active RC because: Active RC: feedback configuration improves linearity; Gm-C filter: open loop operation, the gm stage sees

large signal swing, thus linearization technique is needed, which adds power consumption.

Active RC is preferable for low frequency applications if linearity is a key issue

45 ECEN 622(ESS) Analog and Mixed Signal Center TAMU

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References J. Alvin Connelly and P. Choi,” Macromodeling with Spice”, Prentice Hall, Englewood Cliffs, NJ 07632. 1992

Note that SPICE allows you to describe non-linear components: Nonlinear capacitor and inductors CXXXXXXXX N+ N- POLY C0 C1 C2 …….<IC=INCOND> LYYYYYYYY N+ N- POLY L0 L1 L2 ……..< IC=INCOND> Nonlinear dependent sources E, G, F and H type A two-dimensional polynomial function is expressed as f(x1,x2) = po +p1x1 +p2x2 +p3(x1)*2 + p4x1x2 + p5 (x2)*2 + p6( xi)*3 + p7(x1)*2 x2 + p8x1(x2)*2 + p9(x2)*3 Ename N+ N- <POLY(ndim)> nc1+nc1 -<nc2+ nc2-….> + p0<p1<p2….>> <IC=vnc1,nc1-, vc2+,nc2-,…> E1 3 4 POLY(1) 7 10 10.5 2.0 1.95

Which means VE1 = 10.5 +2.0V7,10 + 1.95(V7,10)*2